R ત્રિજયાના નક્કર ગોળાની અંદર r ત્રિજ્યાનો પો | vedclass

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Question

(b)Let specific gravities of concrete and saw dust are ${\rho _1}$ and ${\rho _2}$ respectively.
According to principle of floatation weight of whole

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(b)Let specific gravities of concrete and saw dust are ${ho _1}$ and ${ho _2}$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3}\pi ({R^3} - {r^3}){ho _1}g + \frac{4}{3}\pi {r^3}{ho _2}g = \frac{4}{3}\pi {R^3} 	imes 1 	imes g$
==> ${R^3}{ho _1} - {r^3}{ho _1} + {r^3}{ho _2} = {R^3}$
==> ${R^3}({ho _1} - 1) = {r^3}({ho _1} - {ho _2})$ ==> $\frac{{{R^3}}}{{{r^3}}} = \frac{{{ho _1} - {ho _2}}}{{{ho _1} - 1}}$
==> $\frac{{{R^3} - {r^3}}}{{{r^3}}} = \frac{{{ho _1} - {ho _2} - {ho _1} + 1}}{{{ho _1} - 1}}$
==> $\frac{{({R^3} - {r^3}){ho _1}}}{{{r^3}{ho _2}}} = \left( {\frac{{1 - {ho _2}}}{{{ho _1} - 1}}} ight)\;\frac{{{ho _1}}}{{{ho _2}}}$
==> $\frac{{{m{Mass of concrete }}}}{{{m{Mass of saw dust}}}} = \left( {\frac{{1 - 0.3}}{{2.4 - 1}}} ight) 	imes \frac{{2.4}}{{0.3}} = 4$

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